The chance of 1 of 2 chances succeeding?

So I have been wondering. A friend told me if you flipped two coins the chance of getting at least 1 tails would still be 50% but i dont agree. I feel like it's higher because you have two coins so you have two chances (50%) to get one tails.
Similarly they told me that if i have a 1/4 chance to get a red marble from a bag that even if i try twice to get a red marble (assuming that you take 1 marble and put it back in before drawing another one out) that i will still have a 25% chance as its 2/8. That doesn't make any sense to me. That's like saying someone who buys 1,000 lotto tickets has the same chance as winning as someone who buys 1. One person has 1,000 chances and another has 1. So even though every individual ticket has the same chance to win the person with more chances has an overall higher chance to win. So whats right here? Thank you in advance to anyone who answers, this has really stumped me!

SerenaWolf said:

So I have been wondering. A friend told me if you flipped two coins the chance of getting at least 1 tails would still be 50% but i dont agree. I feel like it's higher...

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Feelings are nice, but facts and logic matter more for mathematics. What are the four possible results, if you have, say, a fair dime and a fair penny? Each could land "heads" or "tails". What, in list form, are the outcomes? In how many of those outcomes is there "at least one tail"? Is this "3" or "4" (that is, "more than two of four")? Prove your claim!

SerenaWolf said:

Similarly they told me that if i have a 1/4 chance to get a red marble from a bag that even if i try twice to get a red marble (assuming that you take 1 marble and put it back in before drawing another one out)....

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Since we have no information regarding what all was in the bag to start with, we can draw no conclusions.

Well, let's think about the coin flipping exercise. How many outcomes are possible for the first coin? How many outcomes are possible for the second coin? Then, how many outcomes are possible if both are flipped? Of those n outcomes, how many have at least one Tails? What percentage of the total is that? Does that match your intuition, your friend's, or neither?

Similarly with the marble picking exercise. It's a little different, but the same basic process applies. You're given the odds of picking a red marble from the bag, so what's the odds of not picking a red marble? Then you put whatever marble you picked back, so the odds are the same again? Now, again, make a chart of all the possible outcomes just like you did for the coin flipping exercise, being sure to note the probabilities of each branch. The way you worded the question is a bit confusing to me, and I'm not sure if you want to find the chance of picking exactly one red marble, or at least one red marble. Either way, you can use the chart to find your desired answer - just note that they're not the same chance.

Finally, for the lottery exercise, your line of reasoning may or may not be valid. If the person in question bought 1000 lottery tickets, but he picked the same numbers for each ticket, then he would have the same chance to win as someone who bought only 1 ticket. If, however, each of the man's 1000 tickets have unique numbers, then you'd be correct in saying that he stands a greater chance of winning than the man who bought 1 ticket.

SerenaWolf said:

A friend told me if you flipped two coins the chance of getting at least 1 tails would still be 50% but i dont agree. I feel like it's higher because you have two coins so you have two chances (50%) to get one tails.
Similarly they told me that if i have a 1/4 chance to get a red marble from a bag that even if i try twice to get a red marble (assuming that you take 1 marble and put it back in before drawing another one out) that i will still have a 25% chance as its 2/8.

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Here is the outcome space of any binary operation done twice. It could be a coin toss, or a marble drawing.
\(\displaystyle \begin{array}{*{20}{c}}
X&X\\
X&O\\
O&X\\
O&O\end{array}\)

In the coin toss both X & O have 0.5 probability. There is only one row that has both X's.
There are three rows with at least one O. So the probability of at least one O is \(\displaystyle \frac{3}{4}\).

The bag has one red marble and three green marbles. X is for a red and O means a green in drawn.
But this time probabilities are different. Here is the table with the probabilities of each row.
\(\displaystyle \begin{array}{*{20}{c}}
X&X\\
X&O\\
O&X\\
O&O\end{array}\)\(\displaystyle \begin{array}{*{20}{c}}|&{\frac{1}{16} }\\
|&{\frac{3}{16} }\\
|&{\frac{3}{16} }\\
|&{\frac{9}{16} }
\end{array}\)
Add up the rows with at least one X and we get \(\displaystyle \dfrac{7}{16}\) and that is the probability at least one red marble, drawing with replacement.